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how to fit exponential distribution function on data?

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Mos_bad
Mos_bad on 25 Jan 2020
Answered: Image Analyst on 26 Jan 2020
The vector m follows the truncated exponential equation (F_M) and it is shown by solid black line in figure. I intend to fit an exponential distribution function to data and find the parameter lambda (1/mean). Even though I've used fitdist(x,distname), the fitted exp. dist. shown in dashed line which is way different from the data. here is the code:
M_min=4.5; M_max=8.0;
m=M_min:0.0001:M_max;
a=4.56; b=1.0;
alpha=a*log(10);beta=b*log(10);
nu=exp(alpha-beta*M_min);
F_M=(1-exp(-beta*(m-M_min))) / (1-(exp(-beta*(M_max-M_min)))); % CDF of Mag.
pd = fitdist(m','Exponential');
figure(1); plot(m,1-F_M,'-','linewidth',2);
hold on; plot(m,1-cdf(pd,m),'--');
legend('data','fitted dist')

Answers (2)

Walter Roberson
Walter Roberson on 26 Jan 2020
You do not have an exponential distribution. (1 minus an exponential) is not an exponential.
On the other hand if you fit using the equation
a*exp(-b*x)+c
instead of
a*exp(-b*x)
then you get pretty much a perfect fit.

  1 Comment

Mos_bad
Mos_bad on 26 Jan 2020
All I want to do is to devide vertical axis to 1000 intervals and pick a random value of magnitude (horizental axis) at each interval. Kind of Latin hypercube sampling.

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